How do you solve #x^2+4x+7=0#?

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Answer 1 · 2015-08-12T14:45:32

Try factoring first. It won't work, so either use the quadratic formula or complete the square, as you prefer.

#x^2+4x+7=0#

The expression will not factor using integers.

Use your favorite of either the quadratic formula or completing the square.

Completing the square

Because the leading coefficient is #1#, and the coefficient of #x# is even, completing the square is a good choice for this equation.

#x^2+4x " " " " = -7#

#x^2+4x +4 = -7+4#

#(x+2)^2 = -3#

Which has no solution in the Real Numbers, but has 2 imaginary solutions in the Complex Numbers.

#x+2 = +-sqrt(-3) = +-isqrt3#

#x = -2 +- i sqrt3#

Quadratic formula
#x^2+4x+7=0#

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#x = (-(4)+-sqrt((4)^2-4(1)(7)))/(2(1))#

# = (-4+-sqrt(16-28))/2#

# = (-4+-sqrt(-12))/2#

Which has no solution in the Real Numbers, but has 2 imaginary solutions in the Complex Numbers.

# = (-4+- 2isqrt(3))/2#

# = -2 +- isqrt3#